Let \[f(x) = \left\{
\begin{array}{cl}
-x - 3 & \text{if } x \le 1, \\
\frac{x}{2} + 1 & \text{if } x > 1.
\end{array}
\right.\]Find the sum of all values of $x$ such that $f(x) = 0$.
Answer: We solve the equation $f(x) = 0$ on the domains $x \le 1$ and $x > 1.$

If $x \le 1,$ then $f(x) = -x - 3,$ so we want to solve $-x - 3 = 0.$ The solution is $x = -3,$ which satisfies $x \le 1.$

If $x > 1,$ then $f(x) = \frac{x}{2} + 1,$ so we want to solve $\frac{x}{2} + 1 = 0.$ The solution is $x = -2,$ but this value does not satisfy $x > 1.$

Therefore, the only solution is $x = \boxed{-3}.$